Problem 49 - Precalculus Essentials

Question

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$9 \tan ^{2} x-3=0$$

Step-by-Step Solution

Answer

The solution to the equation \(9 \tan^2 x - 3 = 0\), for \(x\) in the interval \([0,2 \pi)\), is \(x = \arctan(\sqrt{1/3}), \arctan(-\sqrt{1/3}), \pi + \arctan(\sqrt{1/3}), \pi + \arctan(-\sqrt{1/3})\).

Step 1: Transform the equation into quadratic form

The given equation is \(9 \tan^2 x - 3 = 0\). It can be rewritten as \(9 \tan^2 x = 3\). Solving for \(\tan^2 x\), we have \(\tan^2 x = 3/9 = 1/3\). So we have transformed the given equation into the quadratic form \(\tan^2 x - 1/3 = 0\).

Step 2: Find the roots of the equation

This equation can be solved by taking the square root on both sides. We get \(\tan x = \sqrt{1/3}\) and \(\tan x = -\sqrt{1/3}\). These are the roots of the given equation.

Step 3: Translate these roots into original variable within the specified range

We solve \(\tan x = \sqrt{1/3}\) and \(\tan x = -\sqrt{1/3}\) in the range \([0,2 \pi)\). We get the solutions as \(x = \arctan(\sqrt{1/3}), \arctan(-\sqrt{1/3}), \pi + \arctan(\sqrt{1/3}), \pi + \arctan(-\sqrt{1/3})\), which fall within the specified range.

Key Concepts

Quadratic Trigonometric Equations

Solving quadratic trigonometric equations is quite similar to solving quadratic algebraic equations, with the distinction that the unknown variable is replaced by a trigonometric function. These equations typically have the form \(a \sin^2(x) + b \sin(x) + c = 0\) or \(a \cos^2(x) + b \cos(x) + c = 0\), although any trigonometric function can apply.

To solve a quadratic trigonometric equation:

• Rewrite the equation in standard quadratic form if necessary.
• Factor the equation, use the quadratic formula, or complete the square to find the solutions for the trigonometric function.
• Translate your solutions into angles, remembering to consider the periodic nature of trigonometric functions.

For our example \(9 \tan ^{2} x-3=0\), we rewrote the equation as \(\tan^2 x = \frac{1}{3}\), which is a simplified quadratic form. By finding the square root of both sides, we identify two tangent values which we then translate back to angles within the given interval using inverse trigonometric functions.

Tan Squared Equations

When dealing with \(\tan^2 x\) equations, such as the one in our exercise, we are essentially tackling the tangent function squared. The squared function indicates that there could be both positive and negative roots upon solving the equation.

In this case, the steps involve:

• Isolating \(\tan^2 x\) to one side of the equation.
• Extracting the square roots, leading to two separate equations: \(+\sqrt{a}\) and \(\-sqrt{a}\).
• Solving each equation for \(x\) within the defined interval.

For \(\tan^2 x = \frac{1}{3}\), this yields \(\tan x = \pm\sqrt{\frac{1}{3}}\), which gives us two equations to solve. After calculating the possible values of \(x\), you should always check if each solution falls within the desired interval, which in this context is \( [0,2 \pi) \).

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find the angle that corresponds to a specific trigonometric function value. Commonly used inverse functions include \(\arcsin\), \(\arccos\), and \(\arctan\), referring to inverse sine, cosine, and tangent functions respectively.

Here are some points to consider when using inverse trigonometric functions:

• They can return principal values, which are the default angles given by calculators or most software programs.
• Since trigonometric functions are periodic, there may be multiple angles that correspond to the same function value.
• Particular attention should be paid to the defined range or domain of the function to ensure all valid angles are considered.

In our example, to solve for \(x\) we use \(\arctan(\sqrt{\frac{1}{3}})\) and \(\arctan(-\sqrt{\frac{1}{3}})\) to find the angles within the interval \( [0,2 \pi) \). It is important to consider the periodicity of the tangent function and ensure solutions take into account all possible angles within the specified range.